HQ5.44 When 0.103 g of Zn(s) combines with enough HCl to make 58.6 mL of HCl(aq)
ID: 588155 • Letter: H
Question
HQ5.44 When 0.103 g of Zn(s) combines with enough HCl to make 58.6 mL of HCl(aq) in a coffee cup calorimeter, all of the zinc reacts, which increases the temperature of the HCl solution from 23.1°C to 24.9"C: Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) Calculate the enthalpy change of the reaction Hrzn in J/mol. Insert your answer in ki, but do not write kJ after the number. (Assume the density of the solution is 1.00 g/mL and the specific heat capacity of solution is 4.184 J/g°C.) Numeric Answer: UnansweredExplanation / Answer
heat released(q) = m*s*DT
m = mass of solution = 58.6+0.103 = 58.703 g
s = specific heat of solution = 4.184 j/g.c
DT = 24.9 - 23.1 = 1.8
q = 58.703*4.184*1.8
= 442.1 joule
no of mol of Zn = 0.103/65.39 = 0.0016 mol
DHrxn = -q/n = -442.1/0.0016 = -276312.5 j/mol
= -276.312 kj/mol
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